Question : AB and CD are two chords in a circle with centre O and AD is a diameter. AB and CD produced meet a point P outside the circle. If $\angle A P D=25^{\circ}$ and $\angle D A P=39^{\circ}$, then the measure of $\angle C B D$ is:
Option 1: 29°
Option 2: 26°
Option 3: 27°
Option 4: 32°
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Correct Answer: 26°
Solution : Given: AB and CD are two chords in a circle with centre O and AD is a diameter AB and CD produced meet at a point P outside the circle $\angle$ APD = 25$^\circ$ $\angle$ DAP = 39$^\circ$ Concept used: Exterior angle theorem - An exterior angle of a triangle is equal to the sum of its two opposite non-adjacent interior angles. The angle subtended from a diameter is 90° Calculation: According to the question We know that an exterior angle of a triangle is equal to the sum of its two opposite non-adjacent interior angles. So, In $\triangle$ ADP $\angle$ ADC = $\angle$ DAP + $\angle$ DPA ⇒ $\angle$ ADC = 25$^\circ$ + 39$^\circ$ ⇒ $\angle$ ADC = 64$^\circ$ Now, Angles in the same segment are equal So, $\angle$ ADC = $\angle$ ABC = 64$^\circ$ ----(1) Since AD is a diameter So, $\angle$ ABD = 90$^\circ$ ⇒ $\angle$ ABC + $\angle$ CBD = 90$^\circ$ ⇒ 64$^\circ$ + $\angle$ CBD = 90$^\circ$ [From (1)] ⇒ $\angle$ CBD = 26$^\circ$ $\therefore$ The value of $\angle$CBD is 26$^\circ$. Hence, the correct answer is 26$^\circ$.
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